Calculate Work to Drain a Hemispherical Tank

[Punkyc7]

a bowl shaped tank is in the shaoe of a hemisphere with a radius of 2 in. If the bowl is filled with water density rho to a depth of 1 in, find the work in pumping the water out fo the tank



W=FD

V=integral of surface area * height

F= rho *V
V is the integral of pi(sqrt(4-y^2))^2 from 0-1
D=2-x

so W=rho*V int 2-x from 0-1

i get 11pi rho g/3 and the answer is suppose to be 9 rho pi/4

im not sure where i am going wrong but i thing it has to do with my integration

 

[HallsofIvy]

Assume that your hemi-spherical bowl has its flat surface in the xy-plane and the center of that surface at (0, 0, 0). Then $x^2+ y^2+ z^2= 4$

Now, imagine a "layer" of water at depth z (z< 0), of thickness "dz". The boundary of that layer is the circle $x^2+ y^2= 4- z^2$ which has radius $\sqrt{4- z^2}$ and so area $\pi(4- z^2)$ and volume $\pi(4- z^2)dz$. The weight of that water is $\pi\rho g(4- z^2)dz$ and we must lift it a height -z (remember that z is negative) out of the bowl. The work done is $-\pi\rho g(4- z^2)dz$.

Integrate that from z= -2 to z= 0.